How to describe a 1.5-dimensional array/tensor?
$begingroup$
Definition: an array is any vector, matrix, or tensor of any order/rank. An $n$-dimensional array is an array whose components may be identified using $n$ indices.
Let $textbf{u}$, $textbf{v}$, and $textbf{w}$ be vectors, and $textbf{X}$ a matrix of arbitrary dimensions.
Define the binary operation $otimes$ according to $left(textbf{u}otimestextbf{v}right)_{i,j}=u_iv_j$, $left(textbf{u}otimestextbf{v}otimestextbf{w}right)_{i,j,k}=u_iv_jw_k$, $left(textbf{X}otimestextbf{u}right)_{i,j,k}=X_{i,j}u_k$ etc - i.e.:
$$textbf{u}otimestextbf{v}=left(begin{matrix}u_1v_1&cdots&u_1v_n\ vdots&ddots&vdots\u_mv_1&cdots&u_mv_nend{matrix}right) textbf{v}otimestextbf{u}=left(begin{matrix}v_1u_1&cdots&v_1u_n\ vdots&ddots&vdots\v_mu_1&cdots&v_mu_nend{matrix}right)$$
$$textbf{u}otimestextbf{v}otimestextbf{w}=left(left(begin{matrix}u_1v_1w_1&cdots&u_1v_nw_1\vdots&ddots&vdots\u_mv_1w_1&cdots&u_mv_nw_1end{matrix}right),cdots,left(begin{matrix}u_1v_1w_l&cdots&u_1v_nw_l\vdots&ddots&vdots\u_mv_1w_l&cdots&u_mv_nw_lend{matrix}right)right)$$
It follows that if $textbf{x}$ is an $m$ dimensional array and $textbf{y}$ is an $n$ dimensional array, then $textbf{x}otimestextbf{y}$ is an $m+n$ dimensional array.
For any array, $textbf{a}$, let $textbf{a}^2=textbf{a}otimestextbf{a}$. It follows that if $textbf{a}$ is an $m$ dimensional object, $textbf{a}^n$ is an $mn$ dimensional object.
For a given array $textbf{X}$, we may define the inverse operation by $textbf{X}^frac{1}{2}=textbf{x}midtextbf{x}^2=textbf{X}$. For example:
$$left(x,yright)^2=left(begin{matrix}x^2&xy\yx&y^2end{matrix}right)impliesleft(begin{matrix}x^2&xy\yx&y^2end{matrix}right)^frac{1}{2}=left(x,yright)$$
Knowing this, define $textbf{U}^x$ for a given array $textbf{U}$ and arbitrary $x$.
To make the problem more tangible, I started with $textbf{U}^{1.5}$, substituting the vector $left(x,yright)$ for $textbf{U}$.
$$1.5=frac{3}{2}implies textbf{U}^{1.5}=left(textbf{U}^3right)^frac{1}{2}$$
$$textbf{U}=left(x,yright)impliestextbf{U}^3=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)$$
$$thereforetextbf{U}^{1.5}=left(x,yright)^{1.5}=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)^frac{1}{2}=Large?$$
linear-algebra abstract-algebra matrices
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
$endgroup$
add a comment |
$begingroup$
Definition: an array is any vector, matrix, or tensor of any order/rank. An $n$-dimensional array is an array whose components may be identified using $n$ indices.
Let $textbf{u}$, $textbf{v}$, and $textbf{w}$ be vectors, and $textbf{X}$ a matrix of arbitrary dimensions.
Define the binary operation $otimes$ according to $left(textbf{u}otimestextbf{v}right)_{i,j}=u_iv_j$, $left(textbf{u}otimestextbf{v}otimestextbf{w}right)_{i,j,k}=u_iv_jw_k$, $left(textbf{X}otimestextbf{u}right)_{i,j,k}=X_{i,j}u_k$ etc - i.e.:
$$textbf{u}otimestextbf{v}=left(begin{matrix}u_1v_1&cdots&u_1v_n\ vdots&ddots&vdots\u_mv_1&cdots&u_mv_nend{matrix}right) textbf{v}otimestextbf{u}=left(begin{matrix}v_1u_1&cdots&v_1u_n\ vdots&ddots&vdots\v_mu_1&cdots&v_mu_nend{matrix}right)$$
$$textbf{u}otimestextbf{v}otimestextbf{w}=left(left(begin{matrix}u_1v_1w_1&cdots&u_1v_nw_1\vdots&ddots&vdots\u_mv_1w_1&cdots&u_mv_nw_1end{matrix}right),cdots,left(begin{matrix}u_1v_1w_l&cdots&u_1v_nw_l\vdots&ddots&vdots\u_mv_1w_l&cdots&u_mv_nw_lend{matrix}right)right)$$
It follows that if $textbf{x}$ is an $m$ dimensional array and $textbf{y}$ is an $n$ dimensional array, then $textbf{x}otimestextbf{y}$ is an $m+n$ dimensional array.
For any array, $textbf{a}$, let $textbf{a}^2=textbf{a}otimestextbf{a}$. It follows that if $textbf{a}$ is an $m$ dimensional object, $textbf{a}^n$ is an $mn$ dimensional object.
For a given array $textbf{X}$, we may define the inverse operation by $textbf{X}^frac{1}{2}=textbf{x}midtextbf{x}^2=textbf{X}$. For example:
$$left(x,yright)^2=left(begin{matrix}x^2&xy\yx&y^2end{matrix}right)impliesleft(begin{matrix}x^2&xy\yx&y^2end{matrix}right)^frac{1}{2}=left(x,yright)$$
Knowing this, define $textbf{U}^x$ for a given array $textbf{U}$ and arbitrary $x$.
To make the problem more tangible, I started with $textbf{U}^{1.5}$, substituting the vector $left(x,yright)$ for $textbf{U}$.
$$1.5=frac{3}{2}implies textbf{U}^{1.5}=left(textbf{U}^3right)^frac{1}{2}$$
$$textbf{U}=left(x,yright)impliestextbf{U}^3=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)$$
$$thereforetextbf{U}^{1.5}=left(x,yright)^{1.5}=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)^frac{1}{2}=Large?$$
linear-algebra abstract-algebra matrices
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
$endgroup$
add a comment |
$begingroup$
Definition: an array is any vector, matrix, or tensor of any order/rank. An $n$-dimensional array is an array whose components may be identified using $n$ indices.
Let $textbf{u}$, $textbf{v}$, and $textbf{w}$ be vectors, and $textbf{X}$ a matrix of arbitrary dimensions.
Define the binary operation $otimes$ according to $left(textbf{u}otimestextbf{v}right)_{i,j}=u_iv_j$, $left(textbf{u}otimestextbf{v}otimestextbf{w}right)_{i,j,k}=u_iv_jw_k$, $left(textbf{X}otimestextbf{u}right)_{i,j,k}=X_{i,j}u_k$ etc - i.e.:
$$textbf{u}otimestextbf{v}=left(begin{matrix}u_1v_1&cdots&u_1v_n\ vdots&ddots&vdots\u_mv_1&cdots&u_mv_nend{matrix}right) textbf{v}otimestextbf{u}=left(begin{matrix}v_1u_1&cdots&v_1u_n\ vdots&ddots&vdots\v_mu_1&cdots&v_mu_nend{matrix}right)$$
$$textbf{u}otimestextbf{v}otimestextbf{w}=left(left(begin{matrix}u_1v_1w_1&cdots&u_1v_nw_1\vdots&ddots&vdots\u_mv_1w_1&cdots&u_mv_nw_1end{matrix}right),cdots,left(begin{matrix}u_1v_1w_l&cdots&u_1v_nw_l\vdots&ddots&vdots\u_mv_1w_l&cdots&u_mv_nw_lend{matrix}right)right)$$
It follows that if $textbf{x}$ is an $m$ dimensional array and $textbf{y}$ is an $n$ dimensional array, then $textbf{x}otimestextbf{y}$ is an $m+n$ dimensional array.
For any array, $textbf{a}$, let $textbf{a}^2=textbf{a}otimestextbf{a}$. It follows that if $textbf{a}$ is an $m$ dimensional object, $textbf{a}^n$ is an $mn$ dimensional object.
For a given array $textbf{X}$, we may define the inverse operation by $textbf{X}^frac{1}{2}=textbf{x}midtextbf{x}^2=textbf{X}$. For example:
$$left(x,yright)^2=left(begin{matrix}x^2&xy\yx&y^2end{matrix}right)impliesleft(begin{matrix}x^2&xy\yx&y^2end{matrix}right)^frac{1}{2}=left(x,yright)$$
Knowing this, define $textbf{U}^x$ for a given array $textbf{U}$ and arbitrary $x$.
To make the problem more tangible, I started with $textbf{U}^{1.5}$, substituting the vector $left(x,yright)$ for $textbf{U}$.
$$1.5=frac{3}{2}implies textbf{U}^{1.5}=left(textbf{U}^3right)^frac{1}{2}$$
$$textbf{U}=left(x,yright)impliestextbf{U}^3=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)$$
$$thereforetextbf{U}^{1.5}=left(x,yright)^{1.5}=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)^frac{1}{2}=Large?$$
linear-algebra abstract-algebra matrices
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
$endgroup$
Definition: an array is any vector, matrix, or tensor of any order/rank. An $n$-dimensional array is an array whose components may be identified using $n$ indices.
Let $textbf{u}$, $textbf{v}$, and $textbf{w}$ be vectors, and $textbf{X}$ a matrix of arbitrary dimensions.
Define the binary operation $otimes$ according to $left(textbf{u}otimestextbf{v}right)_{i,j}=u_iv_j$, $left(textbf{u}otimestextbf{v}otimestextbf{w}right)_{i,j,k}=u_iv_jw_k$, $left(textbf{X}otimestextbf{u}right)_{i,j,k}=X_{i,j}u_k$ etc - i.e.:
$$textbf{u}otimestextbf{v}=left(begin{matrix}u_1v_1&cdots&u_1v_n\ vdots&ddots&vdots\u_mv_1&cdots&u_mv_nend{matrix}right) textbf{v}otimestextbf{u}=left(begin{matrix}v_1u_1&cdots&v_1u_n\ vdots&ddots&vdots\v_mu_1&cdots&v_mu_nend{matrix}right)$$
$$textbf{u}otimestextbf{v}otimestextbf{w}=left(left(begin{matrix}u_1v_1w_1&cdots&u_1v_nw_1\vdots&ddots&vdots\u_mv_1w_1&cdots&u_mv_nw_1end{matrix}right),cdots,left(begin{matrix}u_1v_1w_l&cdots&u_1v_nw_l\vdots&ddots&vdots\u_mv_1w_l&cdots&u_mv_nw_lend{matrix}right)right)$$
It follows that if $textbf{x}$ is an $m$ dimensional array and $textbf{y}$ is an $n$ dimensional array, then $textbf{x}otimestextbf{y}$ is an $m+n$ dimensional array.
For any array, $textbf{a}$, let $textbf{a}^2=textbf{a}otimestextbf{a}$. It follows that if $textbf{a}$ is an $m$ dimensional object, $textbf{a}^n$ is an $mn$ dimensional object.
For a given array $textbf{X}$, we may define the inverse operation by $textbf{X}^frac{1}{2}=textbf{x}midtextbf{x}^2=textbf{X}$. For example:
$$left(x,yright)^2=left(begin{matrix}x^2&xy\yx&y^2end{matrix}right)impliesleft(begin{matrix}x^2&xy\yx&y^2end{matrix}right)^frac{1}{2}=left(x,yright)$$
Knowing this, define $textbf{U}^x$ for a given array $textbf{U}$ and arbitrary $x$.
To make the problem more tangible, I started with $textbf{U}^{1.5}$, substituting the vector $left(x,yright)$ for $textbf{U}$.
$$1.5=frac{3}{2}implies textbf{U}^{1.5}=left(textbf{U}^3right)^frac{1}{2}$$
$$textbf{U}=left(x,yright)impliestextbf{U}^3=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)$$
$$thereforetextbf{U}^{1.5}=left(x,yright)^{1.5}=left(left(begin{matrix}x^3&x^2y\xyx&xy^2end{matrix}right),left(begin{matrix}yx^2&yxy\y^2x&y^3end{matrix}right)right)^frac{1}{2}=Large?$$
linear-algebra abstract-algebra matrices
linear-algebra abstract-algebra matrices
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
asked Dec 6 '18 at 0:14
R. BurtonR. Burton
40419
40419
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're making an assumption when you say "we may define the inverse operation", which is that the operation is invertible. The definition given for an array only makes sense for $ninmathbb{N}$, and since $text{Dim}(X) = text{Dim}(x^2) = 2text{Dim}(x) = 2n$ for some natural $n$, the operation cannot map to arrays of dimension, for example, $3$. Since arrays of dimension $3$ exist, it must be that the operation is not onto, and thus is not invertible.
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
$endgroup$
$begingroup$
I was using "dimension" to refer to the number of indices, not the number members. By this definition, an array of dimension $3$ is simply a tensor with $3$ indices, and can be constructed using appropriate vectors $textbf{u}$,$textbf{v}$, and $textbf{w}$ by $textbf{u}otimestextbf{v}otimestextbf{w}$. The dimension (in this sense) of a matrix composed of two vectors with $n$ members is $2$, not $n$.
$endgroup$
– R. Burton
Dec 6 '18 at 0:35
$begingroup$
Then, in any case, the argument stands, but for even $n$ rather than perfect-square $n$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:38
$begingroup$
The idea is to $extend$ the operation first to the rationals, then to real $n$. The notion of a '1.5' dimensional array isn't something that's well established, it's a logical consequence of treating $otimes$ as a direct analogue of scalar multiplication. An array of dimension $3$ would be the 'square' of an array of dimension $1.5$. The problem isn't demonstrating that this is the case, it's describing what, exactly, a $1.5$ dimensional array is.
$endgroup$
– R. Burton
Dec 6 '18 at 0:45
$begingroup$
You can define a new set of "arrays" of fractional dimension the same way we do with numbers. Define "arrays" as an object with a rational dimension, with a product that adds dimensions consistent with actual arrays for natural dimension, and a square root that halves dimension consistent with actual arrays for even natural dimension. The issue is that these new "arrays" are not consistent with the old definition of arrays (hence the need to extend the set) and so they won't be written the same way. You can describe the set by its definition, but you can't elements down the same way as arrays.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:05
1
$begingroup$
Well... that certainly makes things more difficult.
$endgroup$
– R. Burton
Dec 6 '18 at 1:14
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027862%2fhow-to-describe-a-1-5-dimensional-array-tensor%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're making an assumption when you say "we may define the inverse operation", which is that the operation is invertible. The definition given for an array only makes sense for $ninmathbb{N}$, and since $text{Dim}(X) = text{Dim}(x^2) = 2text{Dim}(x) = 2n$ for some natural $n$, the operation cannot map to arrays of dimension, for example, $3$. Since arrays of dimension $3$ exist, it must be that the operation is not onto, and thus is not invertible.
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
$endgroup$
$begingroup$
I was using "dimension" to refer to the number of indices, not the number members. By this definition, an array of dimension $3$ is simply a tensor with $3$ indices, and can be constructed using appropriate vectors $textbf{u}$,$textbf{v}$, and $textbf{w}$ by $textbf{u}otimestextbf{v}otimestextbf{w}$. The dimension (in this sense) of a matrix composed of two vectors with $n$ members is $2$, not $n$.
$endgroup$
– R. Burton
Dec 6 '18 at 0:35
$begingroup$
Then, in any case, the argument stands, but for even $n$ rather than perfect-square $n$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:38
$begingroup$
The idea is to $extend$ the operation first to the rationals, then to real $n$. The notion of a '1.5' dimensional array isn't something that's well established, it's a logical consequence of treating $otimes$ as a direct analogue of scalar multiplication. An array of dimension $3$ would be the 'square' of an array of dimension $1.5$. The problem isn't demonstrating that this is the case, it's describing what, exactly, a $1.5$ dimensional array is.
$endgroup$
– R. Burton
Dec 6 '18 at 0:45
$begingroup$
You can define a new set of "arrays" of fractional dimension the same way we do with numbers. Define "arrays" as an object with a rational dimension, with a product that adds dimensions consistent with actual arrays for natural dimension, and a square root that halves dimension consistent with actual arrays for even natural dimension. The issue is that these new "arrays" are not consistent with the old definition of arrays (hence the need to extend the set) and so they won't be written the same way. You can describe the set by its definition, but you can't elements down the same way as arrays.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:05
1
$begingroup$
Well... that certainly makes things more difficult.
$endgroup$
– R. Burton
Dec 6 '18 at 1:14
|
show 1 more comment
$begingroup$
You're making an assumption when you say "we may define the inverse operation", which is that the operation is invertible. The definition given for an array only makes sense for $ninmathbb{N}$, and since $text{Dim}(X) = text{Dim}(x^2) = 2text{Dim}(x) = 2n$ for some natural $n$, the operation cannot map to arrays of dimension, for example, $3$. Since arrays of dimension $3$ exist, it must be that the operation is not onto, and thus is not invertible.
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
$endgroup$
$begingroup$
I was using "dimension" to refer to the number of indices, not the number members. By this definition, an array of dimension $3$ is simply a tensor with $3$ indices, and can be constructed using appropriate vectors $textbf{u}$,$textbf{v}$, and $textbf{w}$ by $textbf{u}otimestextbf{v}otimestextbf{w}$. The dimension (in this sense) of a matrix composed of two vectors with $n$ members is $2$, not $n$.
$endgroup$
– R. Burton
Dec 6 '18 at 0:35
$begingroup$
Then, in any case, the argument stands, but for even $n$ rather than perfect-square $n$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:38
$begingroup$
The idea is to $extend$ the operation first to the rationals, then to real $n$. The notion of a '1.5' dimensional array isn't something that's well established, it's a logical consequence of treating $otimes$ as a direct analogue of scalar multiplication. An array of dimension $3$ would be the 'square' of an array of dimension $1.5$. The problem isn't demonstrating that this is the case, it's describing what, exactly, a $1.5$ dimensional array is.
$endgroup$
– R. Burton
Dec 6 '18 at 0:45
$begingroup$
You can define a new set of "arrays" of fractional dimension the same way we do with numbers. Define "arrays" as an object with a rational dimension, with a product that adds dimensions consistent with actual arrays for natural dimension, and a square root that halves dimension consistent with actual arrays for even natural dimension. The issue is that these new "arrays" are not consistent with the old definition of arrays (hence the need to extend the set) and so they won't be written the same way. You can describe the set by its definition, but you can't elements down the same way as arrays.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:05
1
$begingroup$
Well... that certainly makes things more difficult.
$endgroup$
– R. Burton
Dec 6 '18 at 1:14
|
show 1 more comment
$begingroup$
You're making an assumption when you say "we may define the inverse operation", which is that the operation is invertible. The definition given for an array only makes sense for $ninmathbb{N}$, and since $text{Dim}(X) = text{Dim}(x^2) = 2text{Dim}(x) = 2n$ for some natural $n$, the operation cannot map to arrays of dimension, for example, $3$. Since arrays of dimension $3$ exist, it must be that the operation is not onto, and thus is not invertible.
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
$endgroup$
You're making an assumption when you say "we may define the inverse operation", which is that the operation is invertible. The definition given for an array only makes sense for $ninmathbb{N}$, and since $text{Dim}(X) = text{Dim}(x^2) = 2text{Dim}(x) = 2n$ for some natural $n$, the operation cannot map to arrays of dimension, for example, $3$. Since arrays of dimension $3$ exist, it must be that the operation is not onto, and thus is not invertible.
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
edited Dec 6 '18 at 0:37
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
answered Dec 6 '18 at 0:26
AlexanderJ93AlexanderJ93
6,093823
6,093823
$begingroup$
I was using "dimension" to refer to the number of indices, not the number members. By this definition, an array of dimension $3$ is simply a tensor with $3$ indices, and can be constructed using appropriate vectors $textbf{u}$,$textbf{v}$, and $textbf{w}$ by $textbf{u}otimestextbf{v}otimestextbf{w}$. The dimension (in this sense) of a matrix composed of two vectors with $n$ members is $2$, not $n$.
$endgroup$
– R. Burton
Dec 6 '18 at 0:35
$begingroup$
Then, in any case, the argument stands, but for even $n$ rather than perfect-square $n$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:38
$begingroup$
The idea is to $extend$ the operation first to the rationals, then to real $n$. The notion of a '1.5' dimensional array isn't something that's well established, it's a logical consequence of treating $otimes$ as a direct analogue of scalar multiplication. An array of dimension $3$ would be the 'square' of an array of dimension $1.5$. The problem isn't demonstrating that this is the case, it's describing what, exactly, a $1.5$ dimensional array is.
$endgroup$
– R. Burton
Dec 6 '18 at 0:45
$begingroup$
You can define a new set of "arrays" of fractional dimension the same way we do with numbers. Define "arrays" as an object with a rational dimension, with a product that adds dimensions consistent with actual arrays for natural dimension, and a square root that halves dimension consistent with actual arrays for even natural dimension. The issue is that these new "arrays" are not consistent with the old definition of arrays (hence the need to extend the set) and so they won't be written the same way. You can describe the set by its definition, but you can't elements down the same way as arrays.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:05
1
$begingroup$
Well... that certainly makes things more difficult.
$endgroup$
– R. Burton
Dec 6 '18 at 1:14
|
show 1 more comment
$begingroup$
I was using "dimension" to refer to the number of indices, not the number members. By this definition, an array of dimension $3$ is simply a tensor with $3$ indices, and can be constructed using appropriate vectors $textbf{u}$,$textbf{v}$, and $textbf{w}$ by $textbf{u}otimestextbf{v}otimestextbf{w}$. The dimension (in this sense) of a matrix composed of two vectors with $n$ members is $2$, not $n$.
$endgroup$
– R. Burton
Dec 6 '18 at 0:35
$begingroup$
Then, in any case, the argument stands, but for even $n$ rather than perfect-square $n$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:38
$begingroup$
The idea is to $extend$ the operation first to the rationals, then to real $n$. The notion of a '1.5' dimensional array isn't something that's well established, it's a logical consequence of treating $otimes$ as a direct analogue of scalar multiplication. An array of dimension $3$ would be the 'square' of an array of dimension $1.5$. The problem isn't demonstrating that this is the case, it's describing what, exactly, a $1.5$ dimensional array is.
$endgroup$
– R. Burton
Dec 6 '18 at 0:45
$begingroup$
You can define a new set of "arrays" of fractional dimension the same way we do with numbers. Define "arrays" as an object with a rational dimension, with a product that adds dimensions consistent with actual arrays for natural dimension, and a square root that halves dimension consistent with actual arrays for even natural dimension. The issue is that these new "arrays" are not consistent with the old definition of arrays (hence the need to extend the set) and so they won't be written the same way. You can describe the set by its definition, but you can't elements down the same way as arrays.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:05
1
$begingroup$
Well... that certainly makes things more difficult.
$endgroup$
– R. Burton
Dec 6 '18 at 1:14
$begingroup$
I was using "dimension" to refer to the number of indices, not the number members. By this definition, an array of dimension $3$ is simply a tensor with $3$ indices, and can be constructed using appropriate vectors $textbf{u}$,$textbf{v}$, and $textbf{w}$ by $textbf{u}otimestextbf{v}otimestextbf{w}$. The dimension (in this sense) of a matrix composed of two vectors with $n$ members is $2$, not $n$.
$endgroup$
– R. Burton
Dec 6 '18 at 0:35
$begingroup$
I was using "dimension" to refer to the number of indices, not the number members. By this definition, an array of dimension $3$ is simply a tensor with $3$ indices, and can be constructed using appropriate vectors $textbf{u}$,$textbf{v}$, and $textbf{w}$ by $textbf{u}otimestextbf{v}otimestextbf{w}$. The dimension (in this sense) of a matrix composed of two vectors with $n$ members is $2$, not $n$.
$endgroup$
– R. Burton
Dec 6 '18 at 0:35
$begingroup$
Then, in any case, the argument stands, but for even $n$ rather than perfect-square $n$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:38
$begingroup$
Then, in any case, the argument stands, but for even $n$ rather than perfect-square $n$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:38
$begingroup$
The idea is to $extend$ the operation first to the rationals, then to real $n$. The notion of a '1.5' dimensional array isn't something that's well established, it's a logical consequence of treating $otimes$ as a direct analogue of scalar multiplication. An array of dimension $3$ would be the 'square' of an array of dimension $1.5$. The problem isn't demonstrating that this is the case, it's describing what, exactly, a $1.5$ dimensional array is.
$endgroup$
– R. Burton
Dec 6 '18 at 0:45
$begingroup$
The idea is to $extend$ the operation first to the rationals, then to real $n$. The notion of a '1.5' dimensional array isn't something that's well established, it's a logical consequence of treating $otimes$ as a direct analogue of scalar multiplication. An array of dimension $3$ would be the 'square' of an array of dimension $1.5$. The problem isn't demonstrating that this is the case, it's describing what, exactly, a $1.5$ dimensional array is.
$endgroup$
– R. Burton
Dec 6 '18 at 0:45
$begingroup$
You can define a new set of "arrays" of fractional dimension the same way we do with numbers. Define "arrays" as an object with a rational dimension, with a product that adds dimensions consistent with actual arrays for natural dimension, and a square root that halves dimension consistent with actual arrays for even natural dimension. The issue is that these new "arrays" are not consistent with the old definition of arrays (hence the need to extend the set) and so they won't be written the same way. You can describe the set by its definition, but you can't elements down the same way as arrays.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:05
$begingroup$
You can define a new set of "arrays" of fractional dimension the same way we do with numbers. Define "arrays" as an object with a rational dimension, with a product that adds dimensions consistent with actual arrays for natural dimension, and a square root that halves dimension consistent with actual arrays for even natural dimension. The issue is that these new "arrays" are not consistent with the old definition of arrays (hence the need to extend the set) and so they won't be written the same way. You can describe the set by its definition, but you can't elements down the same way as arrays.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 1:05
1
1
$begingroup$
Well... that certainly makes things more difficult.
$endgroup$
– R. Burton
Dec 6 '18 at 1:14
$begingroup$
Well... that certainly makes things more difficult.
$endgroup$
– R. Burton
Dec 6 '18 at 1:14
|
show 1 more comment
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027862%2fhow-to-describe-a-1-5-dimensional-array-tensor%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
